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A single electron orbits around a statio...

A single electron orbits around a stationary nucleus of charge +Ze. It requires 47.2 eV to excite the electron from the `2^"nd"` to `3^"rd"` Bohr orbit. Find atomic number 'Z' of atom .

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To find the atomic number \( Z \) of the atom, we start by using the formula for the energy levels of a hydrogen-like atom: \[ E = -\frac{13.6 Z^2}{n^2} \] where \( E \) is the energy of the electron in the orbit, \( Z \) is the atomic number, and \( n \) is the principal quantum number of the orbit. ### Step 1: Determine the energy difference between the orbits The energy required to excite the electron from the \( n = 2 \) orbit to the \( n = 3 \) orbit is given as \( 47.2 \, \text{eV} \). The energy difference between these two levels can be expressed as: \[ E_{3} - E_{2} = -\frac{13.6 Z^2}{3^2} - \left(-\frac{13.6 Z^2}{2^2}\right) \] ### Step 2: Substitute the values of \( n \) Substituting \( n = 2 \) and \( n = 3 \): \[ E_{3} - E_{2} = -\frac{13.6 Z^2}{9} + \frac{13.6 Z^2}{4} \] ### Step 3: Find a common denominator The common denominator for \( 9 \) and \( 4 \) is \( 36 \): \[ E_{3} - E_{2} = -\frac{13.6 Z^2 \cdot 4}{36} + \frac{13.6 Z^2 \cdot 9}{36} \] This simplifies to: \[ E_{3} - E_{2} = \frac{13.6 Z^2 (9 - 4)}{36} = \frac{13.6 Z^2 \cdot 5}{36} \] ### Step 4: Set the energy difference equal to the given energy Now we set this equal to the energy required for the transition: \[ \frac{13.6 Z^2 \cdot 5}{36} = 47.2 \] ### Step 5: Solve for \( Z^2 \) To isolate \( Z^2 \), we multiply both sides by \( 36 \): \[ 13.6 Z^2 \cdot 5 = 47.2 \cdot 36 \] Calculating the right side: \[ 47.2 \cdot 36 = 1699.2 \] Now, we have: \[ 68 Z^2 = 1699.2 \] Dividing both sides by \( 68 \): \[ Z^2 = \frac{1699.2}{68} \approx 25 \] ### Step 6: Find \( Z \) Taking the square root of both sides gives: \[ Z = \sqrt{25} = 5 \] ### Final Answer The atomic number \( Z \) of the atom is \( 5 \). ---
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